Question: Let $f(x)=-3 x^4-8 x^3-6 x^2+1$. What is the absolute maximum value of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $0$ (Choice C) C $10$ (Choice D) D $f$ has no maximum value
Explanation: Let's first find the relative extremum points of $f$, and then consider them along with the function's end behavior in both directions. We start with finding the critical points of $f$. The derivative of $f$ is $f'(x)=-12x(x+1)^2$. $f'(x)=0$ for $x=-1,0$. $f'$ is defined for all real numbers. Therefore, our critical points are $x=-1$ and $x=0$. Our critical points divide the function's domain (which is all real numbers) into three intervals: $\llap{-}2$ $\llap{-}1$ $0$ $1$ $x< \llap{-}1$ $\llap{-}1<x<0$ $x>0$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $x<-1$ $x=-2$ $f'(-2)=24>0$ $f$ is increasing $\nearrow$ $-1<x<0$ $x=-\dfrac12$ $f'\left(-\dfrac12\right)=\dfrac32>0$ $f$ is increasing $\nearrow$ $x>0$ $x=1$ $f'(1)=-48<0$ $f$ is decreasing $\searrow$ Now let's look at all the critical points: $x$ $f(x)$ Before After Verdict $-1$ $0$ $\nearrow$ $\nearrow$ Not an extremum $0$ $1$ $\nearrow$ $\searrow$ Maximum Let's imagine ourselves walking on the graph of $f$, starting all the way to the left (from $-\infty$ ) and going all the way to the right (until $+\infty$ ). According to the table, we will start by going up until $(0,1)$, and then forever go down. This means that $\lim_{x\to-\infty}f(x)=\lim_{x\to +\infty}f(x)=-\infty$, which means $f$ has no minimum value. However, $f$ does reach an absolute maximum point at $(0,1)$, which means its absolute maximum value is $1$. In conclusion, the absolute maximum value of $f$ is $1$.